Let B _ i (i=1,2,3) be three independent events in a sample space… - Vidyadip

Question

Let $B _{i}(i=1,2,3)$ be three independent events in a sample space. The probability that only $B _{1}$ occur is $\alpha,$ only $B _{2}$ occurs is $\beta$ and only $B _{3}$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B _{i}$ occurs and these $4$ probabilities satisfy the equations $(\alpha-2 \beta) p =\alpha \beta$ and $(\beta-3 \gamma) p =2 \beta \gamma$ (All the probabilities are assumed to lie in the interval $(0,1))$. Then $\frac{ P \left( B _{1}\right)}{ P \left( B _{3}\right)}$ is equal to ..........

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Answer

b
Let $P \left( B _{1}\right)= p _{1}, P \left( B _{2}\right)= p _{2}, P \left( B _{3}\right)= p _{3}$

given that $p _{1}\left(1- p _{2}\right)\left(1- p _{3}\right)=\alpha$ $.....(i)$

$p _{2}\left(1- p _{1}\right)\left(1- p _{3}\right)=\beta$ $....(ii)$

$p _{3}\left(1- p _{1}\right)\left(1- p _{2}\right)=\gamma$ $.......(iii)$

and $\quad\left(1- p _{1}\right)\left(1- p _{2}\right)\left(1- p _{3}\right)= p \ldots \ldots (iv)$

$\Rightarrow \quad \frac{ p _{1}}{1- p _{1}}=\frac{\alpha}{ p }, \frac{ p _{2}}{1- p _{2}}=\frac{\beta}{ p } \& \frac{ p _{3}}{1- p _{3}}=\frac{\gamma}{ p }$

Also $\beta=\frac{\alpha p }{\alpha+2 p }=\frac{3 \gamma p }{ p -2 \gamma}$

$\Rightarrow \alpha p -2 \alpha \gamma=3 \alpha \gamma+6 p \gamma$

$\Rightarrow \quad \alpha p -6 p \gamma=5 \alpha \gamma$

$\Rightarrow \quad \frac{ p _{1}}{1- p _{1}}-\frac{6 p _{3}}{1- p _{3}}=\frac{5 p _{1} p _{3}}{\left(1- p _{1}\right)\left(1- p _{3}\right)}$

$\Rightarrow p _{1}-6 p _{3}=0$

$\Rightarrow \frac{ p _{1}}{ p _{3}}=6$

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